How many times will the energy of the electric field of a flat air capacitor change after increasing

How many times will the energy of the electric field of a flat air capacitor change after increasing the charge on its plates by 3 times and filling the space between them with a dielectric with a dielectric constant of 6.

To calculate the change in the energy of the electric field of the taken flat capacitor, we use the ratio: n = Wk / Wn = (qk ^ 2 / 2Sk) / (qn ^ 2 / 2Sn) = (qk ^ 2 / (εk * ε0 * S / d)) / (qn ^ 2 / (εn * ε0 * S / d)) = εn * qk ^ 2 / (qn ^ 2 * εk).

Variables: qc (final charge on the plates) = 3qn (initial charge); εн – dielectric constant of air (εн = 1); εк – dielectric constant of the used dielectric (εк = 6).

Let’s calculate: n = εn * qk ^ 2 / (qn ^ 2 * εk) = 1 * (3qn) ^ 2 / (qn ^ 2 * 6) = 1 * 9 / (1 * 6) = 1.5 p.

Answer: The energy of the electric field of the taken flat capacitor will increase by 1.5 times.